Homework 2 Solutions

# Homework 2 Solutions - 5.2{a A plot indicates that a single...

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Unformatted text preview: 5.2 {a} A plot indicates that a single real root occurs at about 1' = 0.45. (I3) First iteration: :rr = ﬂ = {15 2 1— El 1+ CI ftojf{o.5) = asters) = —o.4s ea = >< lﬂﬂ‘lﬁ = lﬂﬂ‘li': Therefore. the ne'Ia.r bracket is r; = ﬂ and x“ = [15. The process can be repeated until the approximate error falls belo'la.r 10%. As smnnrarized below. this oeeurs after 5 iterations yielding a root estimate of U.4ﬁ3?5_ iteration Xi Xu Xr f xt f Xr in K I: ea 1 [II 1 +3.5 -2.3 [12 43.45 10D.DD% 2 [II 13.5 13.25 -2.3 b.8625 'l .913 3T5 10D.DD% 3 13.25 13.5 +3.3?5 418525 -D.3?81 [1.25 54138 33.33% 4 ﬂ_3?5 13.5 +3.43% —D.3D?EH 3.051395 [1.015691 14.29% 5 13.43% 13.5 {Litﬁﬂ'tﬁ {1.051398 ELM-4 3TB -{II_£HZII3132 E_ET% 5.3 (a) A plot indicates that a single real root occurs at about 1' = 0.55. 8 4 0 4 5 1 1 5 .3 (b) Bisection: First iteration: x, = “'5 +1: 015 2 an = 1"” xi0055=555555 1+0.5 ft0.5}f(0.i5) = 421071205105} = —5.45055 Therefore, the netsr bracket is I: = 0.5 and I}; = 0.?5. The process can be repeated until the approximate error falls belo‘lar 10%. As summarized below. this occurs after 4 iterations yielding a root estimate of 0. 53125. iteration Ir J: x 1 0.50000 1.00000 0.?5000 —1 .218?5 2.83105 —3.4 5035 33.33% 2 0.50000 0.?5000 0.52500 —1 .218?5 1 18488 —1 .4 5838 20.00% 3 0.50000 0.82500 0.58250 —1.218?5 0.10800 —0.12018 11.1190 4 0.50000 0.58250 0.53125 —1 .218?5 0.52422 0.83888 5.88% (e) False position: First iteration: x,-= 0.5 355,-}: 4.11055 .3]; = 1 = 5 5 0.5—1 5, =1—{—) =0.5si'5rs —l.213?5—5 f{0.5)f(0.59?99) = —l.21?5(0.?505?} = —0.514?5 Therefore, the bracket is 11-: 0.5 and :rﬂ = 0.59299. Second iteration: 5-.-=0.5 ﬂi-;}=—1.313?5 5-” = 0.55155 ﬁn} = 0.1505? 0.25 05 200.5 - 0.59299] — 1 21875 —0.?505? 0.55054—0.59?99 0.55054 3:; =0.59'.I'99— =0550I54 as =‘ 510055 = 0.545155 The process can be repeated mm] the approximate error falls below 5.2%. As smmnarized below. this occurs after 4 iterations yielding a root estimate of 5.55255. iteration X; x“ x; x” x. x. in x Jr,- a. 1 5.5 1 .55555 -1 .215?5 5.55555 5.55555 5.?555? 5.51425 2 5.5 5.55555I —1 .215?5 5 .T 5552 5.55554 5.571524 5.55551 5.551 '11: 3 5.5 5.55554 —1 .215?5 5 .5 T524 5.55?34 5.55515 5.55244 5.555% 4 5.5 5.55?34 -1 .215?5 5.55515 5.55?55 5.55553 5.55554 5.551 ‘11: 5.5 {a} A graph of the ﬁmction indicates a positive real root at approximater X = 1.2. (b) Using bisection. the ﬁrst iteration is 5.5+2 x. = =l.25 hJ Eu = moss = sass 2—5.5 5 2—5. f{5.5}f(1.25) = 441255519251) = —o.sssas Therefore. the root is in the ﬁrst interval and the upper guess is redeﬁned as 1'1, = 1.25. The second iteration is .3- xr = 5.5 +1.25 =ﬂﬁﬁ 2 5— 2 . Eu = Illa?— 5151;155:423555 5.5—.15 f{5.5}f(5.51‘5} = —3.4riss{—1 .134 1 a} = 4.25551 Consequently. the root is in the second interim] and the lower guess is redeﬁned as x: = 5.525. All the iterations are displayed in the following table: 1' x; In xr f x; xr in 3: Ir 4:; 1 5.55555 2.55 555 1 .2555 5 -3.4 T255 5.15257Ir —5.555 T3 2 5.55555 1.25555 5.5 255 5 -3.4 T255 —1 .23413 4.2 5551 42.5551: 3 5.51555 1.25555 1.55255 4.23413 —5.45?5 5.55451 12.5551: Thus. after three iterations. we obtain a root estimate of 1.5525 with an approximate error of 12.55%. (1‘) Using ﬁtlse position, the ﬁrst iteration is r _ T _ 2.9?259t95 —2) '* _ ' —3.4?259—2.o?15s 5119.5} f0 .43 93 5) = —3.4?259(9.?5o?3) = —2.52?9? Therefore. the root is in the ﬁrst interval and the upper guess is redeﬁned as 1;. = 1.43935. The second iteration is =1.43935 9. ?59?3(9.5 —1.43935) Ir = 1.43935 — _ = 1.3?11? -3.4?139—0.T55?3 1 — . 5 . Ea = Mmmb: 133229»; 1.2311? f(9.5}f{1.2?12?) = —3.4a259{o.2sooa} = —o.so312 Consequently. the root is in the ﬁrst interval and the upper guess is redeﬁned as I“ = 1.2?12?. All the iterations are displayed in the following table: iteration It 1m in 1" xi. I: f as f an id an as 'l 9.5 2.99999 —3.4?259 2.9?259 1 .43935 9.?55?9 -2.52?9? 2 9.5 '| .43 935 —3.4?259 9.?55?9 1.2? 12? 9.2999? -9.99312 13.22291: 3 9.5 '|.2?12? —3.4?259 9.2999? 1 .21 ?53 9.99?31 9.39319 4.41491: After three iterations we obtain a root estimate of 1.21?53 with an approximate error of 4.414311. 5.9 A graph of the ﬁmetion indieatee a poeitit-‘e real root at ap:1:1rox_1'11:11atelji.r :c = 3.2. 10 -1 0 Using ﬂake position. the ﬁrst iteration is _111.431a2{:1—5) —5 41142132 fwjfﬂnmua) = —5(—4.22944) = 21.142 r=5 r = 1.02003 Therefore. the root is in the second interval and the lower guests is redeﬁned as 21;: 1.03003. The remaining iterations are gunmmrized helo'la.r 1' If x“. 1a In x, x; 1" X1 10’ Xr :3 '| 0 5.00000 —5 10.43102 1 .02003 41.22044 21 .14?22 2 1.02003 5.00000 41.2204 10.43102 2.5050? 4.?2004 20.004 50 3?.5?3% 3 2.5 050TIr 5.00000 4.?200 10.43102 3.345 32 214210 10.13210 22.42?% 4 3.34532 5.00000 —2.'l422 10.43102 3021'22 -0.0 002? 1 410300 ?.??2% 5 3.02?22 5.00000 -0.0 003 10.43102 3.?1242 0.10200 0.13500 2.20 5% 0 3.? 1242 5.00000 0100 10.43102 3.?3020 -0.05424 0.01000 0 .03 0% The ﬁnal IEﬁull', X, = 3.23 030, can be checked by auhetituting it into the original ﬁmeﬁon to yield a near-zero reqult, f(3.?3623} = {3.2302313 cos 03.23523 — 5 = 41.115424 ...
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## This note was uploaded on 01/20/2011 for the course ENG EAD 115 taught by Professor Rocke during the Spring '10 term at UC Davis.

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Homework 2 Solutions - 5.2{a A plot indicates that a single...

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